Evaluate the indefinite integral #\int(x^2-1)/(x^3-3x+1)dx# using substitution?

Question

I got stuck, check my work please?

Calculating substitute
#u=x^3-3x+1#
#du=(3x^2-3)dx#
#dx=(du)/(3x^2-3)#

Implementing the substitution
#\int(x^2-1)/(x^3-3x+1)dx\rArr\int(x^2-1)/u\times(du)/(3x^2-3)#
#\rArr\int\cancel((x^2-1))/u\times(du)/(3\cancel((x^2-1)))\rArr\color(red)(\int(du)/(3u))#

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Answer 1 · 2017-02-15T01:00:51

The integral equals #1/3ln|x^3 - 3x + 1| + C#

Your work is completely correct. Here is a look at the next steps from where you left off.

#int 1/(3u)du = 1/3int1/udu = 1/3ln|u| + C#

You simply reverse the substitution now.

#=1/3ln|x^3 - 3x + 1| + C#

Hopefully this helps!